How to go from $dx = -t^2dt$ to $\frac{d}{dx} = -t^2\frac{d}{dt}$

Question

Suppose $x = 1/t$. So now $x$ is a function of $t$, i.e., $x(t)$.

So $$\frac{dx(t)}{dt} = -t^{-2} \Rightarrow dx(t) = -t^{-2}dt$$

This problem is from the textbook: advanced mathematical methods for scientists and engineers

How to go from "$dx = -t^2dt$" to "$\frac{d}{dx} = -t^2\frac{d}{dt}$"?

It seems that I just divide the previous term by $1$ and then multiply it by $d$.

However, it seems unrealistic to me; can anyone please explain this carefully to me? More specifically, can I multiply $d$, which is like an operator to me. thanks!

$\frac{d}{dx}=\frac{dt}{dx} \frac{d}{dt}$. The $d$'s themselves shouldn't be thought of as multiplying or dividing. You can think of the whole unit $dx$ or $dt$ as being multiplied/divided, with appropriate caveats. — Ian, Jan 19 '18 at 03:48
@Ian $dt/dx \neq -t^2$ which is what is throwing me off to begin with — David Reed, Jan 19 '18 at 03:50
These two expressions are not equivalent. The second one doesn't even make sense. If you show us how you came to that conclusion, or where you got this from, only then we can answer — Dylan, Jan 19 '18 at 03:56
I’m trying to delete my attempt, I was thinking something incorrectly. Plus, I don’t know how to typeset the $\frac{d}{dx}$ derivative operators. — Isosceles, Jan 19 '18 at 04:05

David Reed · Accepted Answer · 2018-01-19T04:21:41.540

2

You use the inverse-function theorem and the chain rule:

Inverse function theorem says $$\frac{dx}{dt} = -t^{-2} \to \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}}= -t^2 $$

The chain rule says : $$\frac{d}{dx} = \frac{dt}{dx}\frac{d}{dt} = -t^2\frac{d}{dt}$$

edited Jan 19 '18 at 04:21

answered Jan 19 '18 at 04:16

David Reed

3,265

Peter Szilas · Answer 2 · 2018-01-19T12:31:27.547

Correct me if wrong :

$x=x(t)$, differentiable on an interval $I$, and $x'(t)\not=0,$

then the inverse function $t=t(x)$ exists, and is differentiable $x=x(t).$

Let $F(t(x))$ arbitrary, differentiable, then:

$\dfrac{d}{dx} F(t(x)) = \dfrac{d}{dt} F(t)×\dfrac{d}{dx}t(x),$ I.e.,

$[\dfrac{d}{dx}]F(t(x)) = [\dfrac{d}{dx}t(x) \dfrac{d}{dt}]F(t(x))$,

or $\dfrac{d}{dx} = t'(x)\dfrac{d}{dt}$, as an operator equation.

With:

$t'(x) = \dfrac{1}{x'(t)}= \dfrac{1}{-t^{-2}}$:

$\dfrac{d}{dx} = -t^2 \dfrac{d}{dt}$.

How to go from $dx = -t^2dt$ to $\frac{d}{dx} = -t^2\frac{d}{dt}$

2 Answers2